Regarding Electric Energy Savings

8/2/2019 Regarding Electric Energy Savings

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NIST Technical Note 1654

Regarding Electric Energy Savings,

Power Factors, and Carbon

Footprints: A Primer

Martin MisakianThomas L. Nelson

William E. Feero


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NIST Technical Note 1654

Regarding Electric Energy Savings,

Power Factors, and Carbon

Footprints: A Primer

Martin Misakian

Thomas L. Nelson

Electronics and Electrical Engineering Laboratory

National Institute of Standards and Technology

Gaithersburg, MD 20899

William E. Feero

P.O Box 489

Reedsville, PA 17084

October 2009

U.S. Department of CommerceGary Locke, Secretary

National Institute of Standards and TechnologyPatrick D. Gallagher, Deputy Director


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National Institute of Standards and Technology Technical Note 1654

Natl. Inst. Stand. Technol. Technical Note 1654, 9 pages (October 2009)

CODEN: NSPUE2


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Calculations

Mesh analysis for determining currents in circuits is described in many texts on circuitanalyses and is not reviewed here [2, 3]. We use the approach to set up simultaneousequations that are solved to determine circulating currentsI1,I2, andI3 in the separate

loops shown in Figures 1(b) and 1(c). The simultaneous equations that are developed andtheir solutions are presented below. We choose specific values of the circuit elements (R,L, C) to demonstrate how the current from the power distribution line is affected with andwithout the inclusion of the capacitance. Since the voltage remains essentially constant,changes in the current mainly influence the amount of electric energy provided by thepower line. Because the number of equations will be few, they can be solved by thesubstitution method or by using matrices and Cramers Rule.

Mesh analysis for the circuit in Figure 1(b) yields the equations

,0)(

and)sin(

32

32

=++

=

LRL

oLL

ZZIZI

tVZIZI

where the impedancesZL = Li andZR =R; i = (-1)0.5.

Assuming the appliance is a central air conditioner with values forL,R, and Vo of 0.065henries, 10.4ohms, and 325 peak amplitude volts (230 volts rms) respectively, thesolutions for the currents in amperes (A) are

.A)sin(25.31

andAi)sin(263.13A)sin(25.31

3

21

tI

ttII

=

==

The phase difference between the current and voltage waveforms from the power line isdetermined by examining the angle associated with the tangent ofI1 as represented in thecomplex plane, i.e., -23 degrees. Therefore the PF is cos(-23

o) or 0.92. The amplitude of

the current from the power line is [(31.25)2 + (13.26)2]0.5 or ~34A, and the currentthrough the inductor,IL, is justI2 I3 or -13.263sin (t)i A.

Mesh analysis for the circuit that includes the capacitor C [Fig. 1(c)] yields the equations

,0)(

and,0)(

),sin(

32

321

21

=++

=++

=

LRL

LCLC

oCC

ZZIZI

ZIZZIZI

tVZIZI

where the impedanceZC= -i/C.

We assume a value for capacitance that will demonstrate the maximum effect forreducing the current from the power line. That is we choose a value ofCthat will satisfythe resonance condition with the inductor, i.e., C= 1/2L = ~108.2 F.


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Solutions for the currents in Figure 1(c) are

A.)sin(25.31

andA,)isin(257.13A)sin(25.31

A,)sin(25.31Ai)sin(1041.5A)sin(25.31

3

2

3

1

tI

ttI

tttI

=

=

=

The current provided by the power line,I1, is reduced, i.e., the amplitude is reduced to~31.3 A from 34 A. Yet the currentI2 necessary to operate the motor is unchanged. Thesource of what appears to be extra current inI2, -13.257sin(t)i A, can be understoodby first examining the other currents in the circuit [Fig. 1(c)]. The current through theinductor,IL, equals -13.257sin(t)i A and is unchanged. However, there is now a currentthrough the capacitor,IC, equal toI1 -I2 or +13.257sin(t)i A, which is 180 degrees out ofphase with respect to the current in the inductor.

The underlying physics provides an understanding of what is happening with theintroduction of the capacitor. When voltage is applied to the circuit in Figure 1(c), thecapacitor is instantly charged and this charge oscillates between the capacitor andinductor [4] creating the current through CandL indicated by the mesh analysis. Theoscillating charge is the origin of the extra current inI2 when the current from thepower line is reduced. Appendix A shows this oscillatory relationship between thecharge on the capacitor and current through the inductor as a function of time.

For the ideal (lossless) conditions we have assumed, the power line does not need toprovide additional current to charge the capacitor-inductor network. This last point canbe verified by performing a mesh analysis of the circuit in Fig. 1(c) with the resistorremoved. The currents through CandL will be the same (13.257sin(t)i A), but thecurrent from the power line will be ~zero. The zero current from the power line is alsoexpected since the impedance of theL-Ccircuit is ~infinite. In a more realistic treatmentof the problemthere will be some resistive losses in theL-Ccircuit which will becompensated for by additional charging by the power line (see below).

Also very noteworthy is that under the above conditions the PF has increased to nearlyunity from 0.92 because there is essentially phase congruence between the current andvoltage waveforms, i.e., the phase ofI1 is the same as V0 sin(t).

While the above calculations are for very ideal conditions, including the use of amatching capacitor, reductions in power line current will occur even when the capacitoris poorly matched. For example, if the capacitor is 50 F different from the matchedvalue (108.2 F), the amplitude ofI1 will be 31.8 A (versus ~34 A) and the power factorwill be 0.98 (versus 0.92). Similarly, introducing small resistances of the connectingwires will not change the results significantly.


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Discussion

Residential loads that decrease the power factor include refrigerators, air conditioners,and washing machines. Typically these appliances are operated only intermittently, sothat average residential power factors may not be very low. If an energy saving device of

the type we have considered is used to increase the power factor, the utility will not haveto supply as much current when certain electrical appliances are operated. However, thehomeowners electric bill will not decrease because, as noted above, the cost isdetermined in part by the product of current, voltage, and power factor. That is, as thecurrent from the power line is reduced by the introduction of capacitance, the powerfactor is increased and the product,Ix Vx PF, remains essentially the same. The utilitywill save a small amount of money (see discussion of possible savings below) because itwill not have to supply as much current to the residence, will have smaller resistive lossesalong the power line, will not emit as much greenhouse gas, etc. Will the homeownersreduce their carbon footprint, are they only helping the utilities reduce their carbonfootprint, or can both take some credit? In addition, should the homeowner consider the

CO2 generated during fabrication of the capacitive device and the associated paybacktime? We leave the answer to these questions up to the reader.

It should be noted that some utilities may charge a penalty for power factors that areconsidered too low, for example below 0.95 [5]. In such cases there could be somesavings by installing devices of the type that has been considered. For example,industrial users of electric energy with large inductive loads might be penalized if stepsare not taken to increase their power factor.

Possible utility savings. For our central air conditioner example given above, let usestimate approximately how much energy is saved because of lower resistive losses in thedistribution system when the customer installs the capacitor. The utilitys total resistanceas seen looking back into the distribution system from the watt-hour meter is assumed forour rough calculation to be less than 0.05 ohms. The power consumed in the utilitycircuits for operating the air conditioner without the capacitor is about 28.8 watts (Irms

2R),

whereIrms is the root mean square (rms) value ofI1 in Figure 1(b), 24 A. With thecapacitor installed, the rms value ofI1 [Fig. 1(c)] becomes 22.1 A and the powerconsumed by the utility is reduced to about 4.4 watts. Therefore, installation of thecapacitor reduces the energy from the utility by 4.4 watts multiplied by the running timeof the air conditioner. If the air conditioning unit runs for 12 hours each day, the energysavings will be about 52.8 watt-hours per day. At 20 cents per kilowatt-hour, the moneysaved by the utility would be approximately 1 cent per day. Since in most parts of theUnited States air conditioners only operate for less than six months of the year, theutilitys annual savings would be about $1.80 for a single residence.


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Appendix A

The relationship between the charge on the capacitor and the current through the inductoras a function of time can be determined from the results that have been calculated above.We find the charge in coulombs (Q) on the capacitor, QC, as a function of time by

integrating the current through the capacitor,IC, i.e.,

QtdttdtIQ CC )cos(

257.13)(sin257.13 === .

The calculated current through the inductor,IL, is

A.)sin(257.13 tIL =

By plottingboth the charge and inductor current as a function of time on the samecoordinates (Figure A1), one can see that the charge on the capacitor goes to zero as thecurrent reaches a maximum through the inductor, and reaches a maximum charge as thecurrent through the inductor goes to zero.

Acknowledgement

We thank Gerald Fitzpatrick for useful conversations which contributed to thepreparation of this Technical Note.

References

[1] http://energystar.custhelp.com/cgi-

bin/energystar.cfg/php/enduser/std_adp.php?p_faqid=4941&p_created=1204908170

[2] S.L. Oppenheimer, F.R. Hess Jr., and J.P. Borchers, Direct and Alternating Currents,McGraw Hill Co., New York (1973).

[3] R.M. Kerchner and G.F. Corcoran, Alternating Current Circuits, John Wiley & Sons,New York (1960).

[4] D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, John Wiley &Sons, Inc, New York (2005).

[5] DOE Fact Sheet, http://www.eere.energy.gov/; search for Reducing Power FactorCost.
http://energystar.custhelp.com/cgi-bin/energystar.cfg/php/enduser/std_adp.php?p_faqid=4941&p_created=1204908170http://energystar.custhelp.com/cgi-bin/energystar.cfg/php/enduser/std_adp.php?p_faqid=4941&p_created=1204908170http://www.eere.energy.gov/http://www.eere.energy.gov/http://energystar.custhelp.com/cgi-bin/energystar.cfg/php/enduser/std_adp.php?p_faqid=4941&p_created=1204908170http://energystar.custhelp.com/cgi-bin/energystar.cfg/php/enduser/std_adp.php?p_faqid=4941&p_created=1204908170


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Figure 1. Model circuits for (a) simplified motor, (b) circuit without

capacitive device, and (c) circuit with device.

Figure A1. Relation between charge on the capacitor and inductor current.The vertical axis has units of charge and current. Insets are after Ref. [4].

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